Question

The $K{\alpha }_1$lines for Ca, Zn, Zr, and Sn have wavelengths of data-custom-editor="chemistry" $3.36,1.44,0.79,\mathrm{and}0.49\mathrm{\AA }$, respectively. Calculate an approximate wavelength for the Ka lines of (a) V, (b) Ni, (c) Se, (d) Br, (e) Cd, and (f) Sb.

Short answer

The approximate wavelength of elements-

S.no.	Element	Wavelength $\mathrm{\lambda }\mathrm{in}\mathrm{\AA }$
a.	V	2.52
b	Ni	1.66
c.	Se	1.10
d.	Br	1.03
e.	Cd	0.54
f.	Sb	0.47

Step-by-step solution

Step 1. Given information

The wavelength of Ca is 3.36, Zn is 1.44, Zr is 0.79 and Sn is 0.49 for $K{\alpha }_1$lines.

Step 2. Wavelength of elements 

A spreadsheet of elements will be formed first for the conversion of a given wavelength into frequency.

The formula for frequency in terms of wavelength is:

$v=\frac{c}{\lambda }$

The approximate wavelength of elements is calculated by preparing a spreadsheet of elements then plotting the graph atomic number of elements Vs $\sqrt{v}\times {10}^8$. Values for the graph will be obtained from the calculation between the cells of the spreadsheet.

Frequency and atomic number show the linear relationship. After the calculation of frequency, a graph will be plotted between atomic number and $\sqrt{v}\times {10}^8$.

Step 3. Spreadsheet for conversion of wavelength into frequency-

Speed of light (c)	3.00E+08
Conversion of Aointo m	1.00E+10
Elements	Atomic number	Wavelength $\lambda (\stackrel{\mathrm{o}}{\mathrm{A}})$	Frequency	$\sqrt{v}$	data-custom-editor="chemistry" $$\begin{gathered} \mathrm{V}=2.52\stackrel{\mathrm{o}}{\mathrm{A}} \\ \mathrm{Ni}=1.66\stackrel{\mathrm{o}}{\mathrm{A}} \\ \mathrm{Se}=1.10\stackrel{\mathrm{o}}{\mathrm{A}} \\ \mathrm{Br}=1.03\stackrel{\mathrm{o}}{\mathrm{A}} \\ \mathrm{Cd}=0.54\stackrel{\mathrm{o}}{\mathrm{A}} \\ \mathrm{Sb}=0.47\stackrel{\mathrm{o}}{\mathrm{A}} \end{gathered}$$
Ca	20	3.36	8.93E+17	9.45E+08	9.45
Zn	30	1.44	2.08E+18	1.44E+09	14.43
Zr	40	0.79	3.80E+18	1.95E+09	19.49
Sn	50	0.49	6.12E+18	2.47E+09	24.74

Step 4. The graph between atomic number and v×108 

From the above spreadsheet:

$$\begin{gathered} \mathrm{Cell}D5=\left(300000000\right)/\left(\mathrm{C}5/10000000000\right) \\ \mathrm{Cell}E5=\mathrm{SQRT}\left(D5\right) \\ \mathrm{Cell}F5=E5\times 0.00000001 \end{gathered}$$

Now, plot the graph between atomic number and $\sqrt{v}\times {10}^8$, which is shown below:

From the plot, it concludes some unknown quantities which are:

$$\begin{gathered} \mathrm{Slope}=1.963 \\ \mathrm{Intercept}=1.575 \\ \mathrm{Equation}=y=1.963x+1.575 \end{gathered}$$

Step 5. Wavelength for elements 

Now the approximate wavelength for elements V, Ni, Se, Br, Cd, and Sb for kα lines are:

S. No.	Element	Atomic number	$\sqrt{v}$	Wavelength $\lambda \left(\stackrel{\mathrm{o}}{\mathrm{A}}\right)$
a.	V	23	1.09E+09	2.52
b.	Ni	28	1.35 E+09	1.66
c.	Se	34	1.65 E+09	1.10
d.	Br	35	1.70 E+09	1.03
e.	Cd	48	2.37 E+09	0.54
f.	Sb	51	2.52 E+09	0.47

localid="1646902886830" $$\begin{gathered} \mathrm{Cell}C39=\left(\left(B23-B32\right)/B31\right)\times 10000000000 \\ \mathrm{Cell}D39=\left(300000000\times 10000000000\right)/\left(C39\times C39\right) \end{gathered}$$